Learn how to solve the inverse of the sinc function using MATLAB

Help matlab how to solve the inverse function of the sinc function

for k=1:length(y)

f=@(x)y(k)*x-sin(x);

ezplot(f);%Draw a graph and observe that the zero point of the function is near x0(k)

z(k)=fzero(f,x0(k));% call the fzero function to find the zero point

endsxf2012 (Contact ta on the site)%% Take y as a data as an example, assuming that the y value is y0, then

%f=@(x)y0-sin(x)/x；%%Use command: %ezplot(f);

%%Draw a picture, observe the function, and find a coordinate x0 near the zero point

%% Then, the desired zero point is

%z=fzero(f,x0);% call the fzero function to find the zero point

%For example, y0=0.6, by using

f=@(x)0.6-sin(x)/x;%drawingezplot(f)hold onplot(,,'r')

%Observe that the zero point is near -2 and 2, use

z1=fzero(f,-2)

%The calculated zero point is x=-1.66

z2=fzero(f,2)

%The calculated zero point is x=1.66

This is the graph of y=sinx/x. In my case, the value of y is known, and I need to get all the values ​​of x, that is, I want to get it through the inverse function. But the problem is that when y=1, x is one value, but y=0.8 is two values, and y=0.1 is many values.

This is the graph of y=sinx/x. In my case, the value of y is known, and I need to get all the values ​​of x, that is, I want to get it through the inverse function. But... on the interval you drew, the function is not monotonic, so its inverse function does not exist, or it is a multi-valued function.

How to create an inverse function in matlab

finv

F inverse cumulative distribution function

Syntax

X = finv(P,V1,V2)

Description

X = finv(P,V1,V2) computes the inverse of the F cdf with numerator degrees of freedom V1 and denominator degrees of freedom V2 for the corresponding probabilities in P. P, V1, and V2 can be vectors, matrices , or multidimensional arrays that all have the same size. A scalar input is expanded to a constant array with the same dimensions as the other inputs.

The parameters in V1 and V2 must all be positive integers, and the values ​​in P must lie on the interval [0 1].

The F inverse function is defined in terms of the F cdf as

where

Examples

Find a value that should exceed 95% of the samples from an F distribution with 5 degrees of freedom in the numerator and 10 degrees of freedom in the denominator.

x = finv(0.95,5,10)

x =

3.3258

You would observe values ​​greater than 3.3258 only 5% of the time by chance.

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